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SUPER 15 UP - TGT Math Test 122
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SUPER 15 UP - TGT Math Test 122
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  • Question 1/15
    1 / -0.25

    Arun has 10% less coins of different countries than Varun has. Varun has 15% more such coins than Charu has. By what percent the number of coins which Charu has is less number of coins which Arun has? (correct to one decimal place)

    Solutions

    Let Charu has x coins

    Number of coins which Varun has = = =

    Number of coins which Arun has = = =

    Therefore, required percentage = = = 3.38%

    Hence, option C is correct.

  • Question 2/15
    1 / -0.25

    What is the least positive value of A in the number 34278A2597, which is divisible by 9?

    Solutions

    Divisibility of 9:- Sum of digits of a number is multiple of 9 or divisible by 9 then the number is divisible by 9.

    Sum of digits of 34278A2597 = 3 + 4 + 2 + 7 + 8 + A + 2 + 5 + 9 + 7 = 47 + A

    Multiple of 9 which is greater than 47 is 54

    So,

    47 + A = 54

    A = 54 – 47 = 7

  • Question 3/15
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    If x = 9, then find the value of:  x− 10x+ 11x− 10x+ 10x + 7
    Solutions

    x5 - 10x4 + 11x3 - 10x2 + 10x + 7

    = x5 - 9x4 - x4 + 9x3 + 2x3 - 10x2 + 10x + 7

    =x5 - x5 - x4 + x4 + 2x3 - 9x2 - x2 + 9x + x + 7

    =2x3 - x3 - x2 + x2 + x + 7

    = x3 + x + 7

    = 729 + 9 + 7

    = 745.

  • Question 4/15
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    Rs 6000 becomes Rs 8340 in K years at rate of simple interest. If rate of interest is 13% per annum, then what is the value of K?
    Solutions

    Given Amount= Rs 8340

    Principal=Rs 6000

    Interest = Amount- Principal=Rs (8340-6000)

    = Rs 2340

    Interest=

    2340 =

    K=3

  • Question 5/15
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    AB is a chord in the minor segment of a circle with centre O. C is a point between A and B on the minor arc AB. The tangents to the circle at A and B meet at the point D. If , then the measure of  is.
    Solutions

    Given,

    ACB = 116°

    So, angle made by chord AB on major arc = 180° – 116° = 64°

    So, AOB = 2 × 64° = 128°

    Now, DAO = DBO = 90° (angle made by tangents)

    In quadrilateral DAOB

    DAO + AOB + OBD + BDA = 360°

    90° + 128° + 90° + BDA = 360°

    BDA = 360° – 308°

    BDA = 52°

    Hence, the correct answer is option D

  • Question 6/15
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    The space diagonal of a cube measures cm. What is the volume of the cube?
    Solutions

    We know that:

    Diagonal of a cube =

    8  =

    Side = 8 cm

    Volume of the cube =

    =

    = 512 cm3

    Hence, option C is correct.

  • Question 7/15
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    If the curved surface area of a cylinder is numerically equal to the perimeter of the square made of length of a wire which is in the form of a circle of radius of 14 cm then find the ratio of sum of the height and radius of the cylinder to the difference of the height and radius of the cylinder, if the height of the cylinder is 2 cm?
    Solutions

    A.T.Q.:

    Perimeter of circle will be equal to the perimeter of the square.

    ⇒ 2×π×R = 4a (R = radius of the circle and a = side of square)

    ⇒ 2× (22/7) × 14 = 4 ×a

    ⇒ a = 22 cm.

    Now, 

    ⇒ 2 × π × r × h = 4 × 22 (r & h = radius and height of the cylinder)

    ⇒ 2 × (22/7) × r × 2 = 88
    ⇒ r = 7 cm

    Now, Required ratio is (7 + 2) : (7 − 2) = 9:5

  • Question 8/15
    1 / -0.25

    How many of the following numbers are divisible by 132?

    660, 754, 924, 1452, 1526, 1980, 2045 and 2170

    Solutions

    Factors of 132 = 2 × 2 × 3 × 11

    A number will be divisible by 132 if it is divisible by 2, 3 and 11.

    From the given number,

    Divisibility of 2 - A number is divisible by 2 if the last digit are 0,2 4, 6, 8.

    Out of the given numbers 660, 754, 924, 1452, 1526, 1980, and 2170 are divisible by 2.

    Divisibility of 3 - A number is divisible by 3 if the sum of the digits of the number is divisible by 3.

    Out of the remaining numbers 660, 924, 1452, and 1980 are divisible by 3.

    Now, all these numbers are divisible by 11.

    Hence, 660, 924, 1452, and 1980 are divisible by 132.

    Hence, option D is the correct answer.

  • Question 9/15
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    A glass jar contains 6 white, 8 black, 4 red and 3 blue marbles. If a single marble is

    chosen at random from the jar, what is the probability that it is black or blue?

    Solutions

    Total number of marbles = 6 + 8 + 4 + 3 = 21

    Favourable outcomes = 8 + 3 = 11

    Probability = favourable outcome/total outcome

    =

    "Hence, option B is the correct answer."

  • Question 10/15
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    (−12) [11 + {7 × (−3)}] ÷ [4 {13 − (−3) × (−6)}] =?
    Solutions

    (−12) [11 + {7 × (−3)}] ÷ [4 {13 − (−3) × (−6)}]

    = (−12) [11 + {−21}] ÷ [4 {13 − 18}]

    = (−12) [−10] ÷ [4 × {−5}]

    = (−12) [−10] ÷ [4 × {−5}]

    = 120 ÷ [−20]

    = −6

  • Question 11/15
    1 / -0.25

    In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. A student attempted all the 200 questions and scored 200 marks. The number of questions, he answered correctly was: 
    Solutions
    Let he answered x questions correctly. 

    Then, number of questions he solved incorrectly = 200 − x 

    As per question, 

    Marks earned by correct attempts − marks deducted by incorrect attempts = Total Marks

    ⇒ 4x − 1×(200 − x) = 200

    ⇒ 5x = 400

    ⇒ x = 80 

    Hence, he answered 80 questions correctly.
  • Question 12/15
    1 / -0.25

    The given table shows the ranking of ten students in two subjects mathematics and statistics.

    The coefficient of rank correlation is:

    Solutions

    Ranking of ten students in two subjects mathematics and statistics:

    Coefficient of rank correlation

    R =

    = 1 – 1.3 = -0.3

  • Question 13/15
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    A dog was sold for ₹ 10,000 at a profit of 25%. For what price should he have sold to gain a 40% profit?
    Solutions

    Selling price at 25% profit = ₹ 10,000

    Cost price =

    =  = ₹ 8,000

    Selling price at profit of 40% = 8000 + 40% of 8000

    =

    = 8000 + 3200 = ₹ 11,200

  • Question 14/15
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    X alone can complete a work in 18 days and Y alone can do the same work in half time of the time taken by X. Working together what part of the same work they can complete in a day?
    Solutions

    Time taken by Y to complete the work = 18/2 = 9 days

    Let the total work be 18 units(LCM of 18 and 9).

    Efficiency of X = 18/18 = 1 unit/day

    Efficiency of Y = 18/9 = 2 units/day

    Work done by X and Y in 1 day = 1 + 2 = 3 units

    Required part = 3/18 = 1/6 Part

    "Hence, option D is the correct answer."

  • Question 15/15
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    What is the largest two-digit number which when divided by 6 and 5 leaves remainder 1 in each case?
    Solutions

    LCM (6, 5) = 30

    The required number will be given by = 30K + 1 

    Here, 30K + 1 < 100 …….( as we require largest two digit number )

    Putting K = 3,

    30×3 + 1 = 91 

    Hence the number is 91.

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